Given that $x^n - \frac1{x^n}$ is expressible as a polynomial in $x - \frac1x$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P\left(x-\frac1x\right) = x^5 - \frac1{x^5}.$
you want to try that without TeX?
x^n - 1/x^n = p(x - 1/x)
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where does z come in?
bruh... aops problem! don't reply to this
To express $x^5 - \frac{1}{x^5}$ as a polynomial in $x - \frac{1}{x}$, we can:
1) Find the expansion of $(x - \frac{1}{x})^5$.
2) Use the expansion to express $x^5 - \frac{1}{x^5}$ in terms of $(x - \frac{1}{x})$.
Let's proceed step by step:
1) Finding the expansion of $(x - \frac{1}{x})^5$:
Using the binomial theorem, we know that
$$(a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5.$$
For $(x - \frac{1}{x})^5$, we have $a = x$ and $b = -\frac{1}{x}$, so:
$$(x - \frac{1}{x})^5 = x^5 + 5x^4(-\frac{1}{x}) + 10x^3(-\frac{1}{x})^2 + 10x^2(-\frac{1}{x})^3 + 5x(-\frac{1}{x})^4 + (-\frac{1}{x})^5.$$
Simplifying, we get:
$$(x - \frac{1}{x})^5 = x^5 - 5x^3 + 5x - \frac{1}{x^5}.$$
2) Expressing $x^5 - \frac{1}{x^5}$ in terms of $(x - \frac{1}{x})$:
Rearranging the equation from step 1, we get:
$$x^5 - \frac{1}{x^5} = (x - \frac{1}{x})^5 + 5x^3 - 5x + \frac{1}{x^5}.$$
Now we have expressed $x^5 - \frac{1}{x^5}$ as a polynomial in $x - \frac{1}{x}$ with real coefficients. Therefore, we can let $P(z) = z^5 + 5z^3 - 5z$.
Thus, $P\left(x - \frac{1}{x}\right) = x^5 - \frac{1}{x^5}.$
To find the polynomial $P(z)$, we need to express $x^5 - \frac{1}{x^5}$ as a polynomial in $x - \frac{1}{x}$.
Let's first simplify $x^5 - \frac{1}{x^5}$. Notice that $x - \frac{1}{x}$ appears to be a good candidate for a substitution.
Let $z = x - \frac{1}{x}$. Then we want to express $x^5 - \frac{1}{x^5}$ in terms of $z$. We can rewrite $z$ as $x - \frac{1}{x} = \frac{x^2 - 1}{x}$.
Now let's express $x^5 - \frac{1}{x^5}$ in terms of $z$. We'll use the binomial expansion formula:
$(x - \frac{1}{x})^5 = x^5 - 5x^3(\frac{1}{x})^2 + 10x(\frac{1}{x})^3 - \frac{1}{x^5}$.
Simplifying this gives:
$(x - \frac{1}{x})^5 = x^5 - 5x^3 + 10x(\frac{1}{x})^3 - \frac{1}{x^5}$.
Since we are given that $x^5 - \frac{1}{x^5}$ is equal to this expression, we can assign $P(z) = x^5 - 5x^3 + 10x(\frac{1}{x})^3 - \frac{1}{x^5}$.
So, the polynomial $P(z)$ that satisfies the condition is $P(z) = z^5 - 5z^3 - 10z - 5$.